Question

A company that manufactures large LCD screens knows that not all pixels on their screen light, even if they spend great care when making them. They know that the manufacturing process has a blank pixel rate, on average, of 5.8 blank pixels in a sheet 8ft by 8ft that will be cut into smaller screens. They believe that the occurrences of blank pixels are independent. Their warranty policy states that they will replace any screen sold that shows more than 2 blank pixels. Suppose the number of blank pixels can be modeled by a Poisson distribution. Complete parts a) through d) below.

a) What is the mean number of blank pixels per square foot?
b) What is the standard deviation of blank pixels per square foot?
c) What is the probability that a 2ft by 3ft screen will have at least one defect?
d) What is the probability that a 2ft by 3ft screen will be replaced because it has too many defects?

Answer 1

The mean number of blank pixels per square foot is 0.09. The standard deviation is 0.3. The probability of at least one defect in a 2ft by 3ft screen is 0.0848. The probability of a 2ft by 3ft screen being replaced because of too many defects can be calculated using the Poisson distribution.

Step-by-step explanation:

a) To find the mean number of blank pixels per square foot, we need to convert the given rate of 5.8 blank pixels in a sheet 8ft by 8ft to the rate per square foot. Since the given sheet has an area of 64 square feet (8ft by 8ft), the rate per square foot is 5.8/64 = 0.09 blank pixels per square foot.

b) The standard deviation of a Poisson distribution is equal to the square root of the mean. So, the standard deviation of blank pixels per square foot is √0.09 = 0.3.

c) To find the probability that a 2ft by 3ft screen will have at least one defect, we need to find the probability that there is no defect and subtract it from 1. The mean number of blank pixels per square foot is 0.09, so the probability of no defect in a square foot is e^(-0.09) = 0.9152. Therefore, the probability of at least one defect in a 2ft by 3ft screen is 1 - 0.9152 = 0.0848.

d) The probability that a 2ft by 3ft screen will be replaced because it has too many defects (more than 2 blank pixels) can be calculated using the Poisson distribution. The probability of having 0, 1, or 2 defects in a square foot is given by: P(X = 0) = e^(-0.09), P(X = 1) = (0.09)e^(-0.09), P(X = 2) = (0.09^2 / 2!)e^(-0.09). The probability of more than 2 defects is equal to 1 minus the sum of these probabilities. So the probability of a 2ft by 3ft screen being replaced is 1 - (P(X = 0) + P(X = 1) + P(X = 2)).